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I'm studying the adiabatic theorem and there is a derivation of a geometric phase factor which incorporates terms of the form $\langle \dot \psi_n (t) | \psi_n (t)\rangle$ where the $\psi_n(t)$ are orthonormal.

It seems clear to me that this should be zero. Differentiating the normalization condition $\langle \psi_n (t) | \psi_n (t)\rangle = 1$ (preserved with unitary evolution) causes the RHS to vanish. So zero for $\langle \dot \psi_n (t) | \psi_n (t)\rangle$ and I'm at a loss as to how to understand the geometric phase which is heavily involved with these terms.

Does anyone know where I'm going wrong and why the geometric phase is not zero simply from differentiating the normalization condition?

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  • $\begingroup$ The product rule is $d/dt \langle \psi | \psi \rangle = \langle \dot{\psi} | \psi \rangle + \langle \psi | \dot{\psi} \rangle = 2Re(\langle \dot{\psi}|\psi\rangle)$. $\endgroup$ – jacob1729 Jun 18 at 10:53
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You are forgetting that both the bra and the ket have a $t$ dependence. We can differentiate $\langle \psi(t)|\psi(t)\rangle$ using the product rule \begin{align} \frac{d}{dt}\langle \psi(t)|\psi(t)\rangle &= \langle\dot{\psi}(t)|\psi(t)\rangle + \langle \psi(t)| \dot{\psi}(t)\rangle\\ &= 0 \end{align} therefore we can say \begin{align} \langle\dot{\psi}(t)|\psi(t)\rangle &= -\langle \psi(t)| \dot{\psi}(t)\rangle\\ &= -\langle\dot{\psi}(t)|\psi(t)\rangle^*\;. \end{align} This implies that the real part of $\langle\dot{\psi}(t)|\psi(t)\rangle$ must be $0$, but the imaginary part can be non-zero.

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  • $\begingroup$ I knew it would be something basic! Thank you. :) $\endgroup$ – JamToast Jun 18 at 11:00

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